The system of equations `lambdax+y+z=0,-x+lambday+z=0,-x-y+lambdaz=0` will have a non-zero solution if real values of `lambda` are given by

Options

A. `lambda=1`
B. `lambda=0`
C. `lambda=-1`
D. none of these

Answer

To determine the values of \( \lambda \) for which the system of equations has a non-zero solution, we can use the concept of determinants. The given system of equations is:

1. \( \lambda x + y + z = 0 \)
2. \( -x + \lambda y + z = 0 \)
3. \( -x - y + \lambda z = 0 \)

We can represent this system in matrix form as \( A \mathbf{v} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{v} \) is the vector of variables \( (x, y, z) \).

The coefficient matrix \( A \) is given by:

\[ A = \begin{bmatrix} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{bmatrix} \]

For the system to have a non-zero solution, the determinant of the matrix \( A \) must be zero:

\[ \text{det}(A) = 0 \]

Now, let's calculate the determinant of \( A \):

\[ \text{det}(A) = \begin{vmatrix} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix, we have:

\[ \text{det}(A) = \lambda \begin{vmatrix} \lambda & 1 \\ -1 & \lambda \end{vmatrix} - 1 \begin{vmatrix} -1 & 1 \\ -1 & \lambda \end{vmatrix} + 1 \begin{vmatrix} -1 & \lambda \\ -1 & -1 \end{vmatrix} \]

Calculating each of the \( 2 \times 2 \) determinants:

1. \( \begin{vmatrix} \lambda & 1 \\ -1 & \lambda \end{vmatrix} = \lambda^2 + 1 \)

2. \( \begin{vmatrix} -1 & 1 \\ -1 & \lambda \end{vmatrix} = -\lambda + 1 \)

3. \( \begin{vmatrix} -1 & \lambda \\ -1 & -1 \end{vmatrix} = 1 - \lambda \)

Substituting these back into the determinant expression:

\[ \text{det}(A) = \lambda(\lambda^2 + 1) - 1(-\lambda + 1) + 1(1 - \lambda) \]

Simplifying this:

\[ = \lambda^3 + \lambda - \lambda + 1 + 1 - \lambda \]
\[ = \lambda^3 - \lambda + 2 \]

Setting the determinant equal to zero:

\[ \lambda^3 - \lambda + 2 = 0 \]

Now, we need to find the values of \( \lambda \) that satisfy this equation. This is a cubic equation, and we can use methods such as synthetic division or numerical methods to find the roots.

After testing possible rational roots (like \( \lambda = -1, 0, 1, 2 \)), we find that \( \lambda = -1 \) is a root. We can factor the cubic polynomial as follows:

\[ \lambda^3 - \lambda + 2 = (\lambda + 1)(\lambda^2 - \lambda + 2) \]

Next, we can analyze the quadratic \( \lambda^2 - \lambda + 2 \). The discriminant \( D \) of this quadratic is:

\[ D = (-1)^2 - 4 \cdot 1 \cdot 2 = 1 - 8 = -7 \]

Since the discriminant is negative, the quadratic has no real roots. Therefore, the only real solution for \( \lambda \) is:

\[ \lambda = -1 \]

Thus, the values of \( \lambda \) for which the system has a non-zero solution is:

\[ \lambda = -1 \]